5 Answers
5

Since $\frac{15x-7}{5}$ is required to be an integer, it is of the form $\frac{2+5j}{15}$ for $j$ an arbitrary integer. Thus, the equation to solve becomes
$$
\left\lfloor\frac{6\big(\frac{2+5j}{15}\big)+5}{8}\right\rfloor =
\frac{15\big(\frac{2+5j}{15}\big)-7}{5}\qquad\text{which simplifies to}\quad
\left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor = j-1
$$
Now observe that the function on the right hand side increases quicker
than the function on the left hand side. So we can solve the question
by finding a maximal solution to the inequality $$
j-1\leq\left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor
$$

A warning: But here we should be careful, because a maximal solution to this inequality is not necessarily a solution to our equality. Also, there might be more solutions than just the maximal solution $j_0$ to this inequality. But since the left hand side and the right hand side are linear in $j$, the set of solutions forms an interval in $\mathbb{Z}$. This means that to give all the solutions, it suffices to find the largest integer below $j_0$ for which the inequality is strict.

Note that for any $n\in\mathbb{Z}$ and $a\in\mathbb{R}$, the
inequality $n\leq\lfloor a\rfloor$ holds if and only if $n\leq a$
holds. This helps, since now we can continue to compute
$$
\begin{split} j-1 \leq \left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor & \Leftrightarrow j-1 \leq \frac{\frac{4}{5}+2j+5}{8}\\
& \Leftrightarrow 8j-8\leq \frac{4}{5}+2j+5\\
& \Leftrightarrow 10j\leq 23\\
& \Leftrightarrow j\leq 2. \end{split}
$$
Therefore, we get the maximal solution for the equation by substituting $j=2$ in $x=\frac{2+5j}{15}$, i.e. $x=\frac{4}{5}$. This is indeed a solution to the equality. As we have noted before, $\frac{4}{5}$ is not necessarily the only solution. Indeed, it is easy to check that $j=1$ also gives a solution to the equation. But for $j\leq 0$ the inequality is strict, so the only solutions to the equation are $x=\frac{4}{5}$ and $x=\frac{7}{15}$.

Can you please explain how you got to this form? $$\frac{2+5j}{15}+k $$
–
Grozav Alex IoanMay 24 '12 at 18:23

1

@GrozavAlexIoan: For $\frac{15x-7}{5}$ to be an integer, it must be that $15x-7$ is divixible by $5$. Or equivalently, that $15x$ has residue $2$ when dividing by $2$. This means that the fractional part of $x$ must be $\frac{2}{15}$, $\frac{7}{15}$ or $\frac{12}{15}$. I summarized that by saying that $x=\frac{2+5j}{15}+k$ for $j=0,1,2$.
–
EgbertMay 24 '12 at 22:35

I see just now that I could just have written $\frac{2+5j}{15}$ and allowing all the $j$, instead of writing $\frac{2+5j}{15}+k$ and only allowing $j=0,1,2$. I'll rewrite my solution
–
EgbertMay 25 '12 at 16:04

I simplified the proof. If anything is unclear: please ask.
–
EgbertMay 25 '12 at 16:27

If you really want to solve it without using the fact that it's multiple-choice, notice that the fractional part of $\frac{6x+5}{8}$ is $Q=\frac{6x+5}{8}-\frac{15x-7}{5}$. Since $Q$ is a fractional part, $0 \leq Q < 1$; you should be able to work backwards to find bounds on $x$. Once you have those bounds, Ilya's observation that $3x-7/5$ is an integer means there are only finitely many values of $x$ you have to worry about checking, so just try them all (or write a program to do it for you).